Is this possible without an initial condition? If so can you show it, I'm trying it myself but I get stuck when having to inverse the Laplace transform
You would put in initial conditions as variables. Like letting y(0) equal w, and letting y'(0)=v. Then, you can pick any initial conditions to see the solution in action.
Great videos. but on this you lost me on the last inverse transform ( blue, right hand side). I've never seen this trick of splitting the squared into a 1+1. If someone could explain this or point me in the right direction it would be appreciated.
in my class, the teacher taught us to assume that y=e^(rx) and solve a quadratic equation. If the r doesn't have any real roots, just use cosine and sine.
That's one way of solving this kind of diff EQ. There is usually more than one way to solve problems in math, so your way isn't wrong per say. The question could ask you to solve the equation using the Laplace transformation, so you'd need to know how to do this as well.
The Laplace transform is particularly useful when you DON'T have the equation equal to zero, but instead there is a forcing function of t, on the other side of the equals sign. The assume y=e^(r*t), and solve a quadratic equation works best for the homogeneous cases. He's using this as an example to demonstrate the concept of solving the same problem via a Laplace transform.
Bottom left of the board, he factored the negative out so he could manipulate the insides of the function. When he took the inverse laplace, he distributed the negative again, making the 3 positive.
Hey, I really appreciate your videos. They inspire me a lot ! I also have a request, can you explain the integral of (e^x*cosx) ? It's a partial integral which I don't know how to solve.
To solve an equation with non-initial conditions instead of initial conditions, you cannot do it directly with the Laplace transform. The Laplace transform only works with initial conditions directly. However, there are other methods we could use. One method is to assign arbitrary initial conditions at t=0 as placeholders, and then solve for them later to match the given information, such as y(0) = u and y'(0) = v. Another method with the way you've given the information, is to simply assign capital T, such that capital T is zero when t=k. This means that T = t - k. Then our delayed start conditions, end up being initial conditions after all, and we can proceed with solving the problem in the capital T domain. I'll use capital S to reflect the difference. And then undo the shift, to get back to the little t domain. For this example: y" + 6*y' + 9*y = 0 y(k) = n, y'(k) = m S^2*Y(S) - S*n - m + 6*S*Y(S) - 6*n + 9*Y(S) = 0 Shuffle initial conditions to the right: (S^2 + 6*S + 9)*Y(S) = S*n + 6*n + m Isolate Y(S) and factor the bottom: Y(S) = (S*n + 6*n + m)/(S + 3)^2 Partial fractions: Y(S) = n/(S+3) + (3*n + m)/(S + 3)^2 Solution in capital T world: y(T) = n*e^(-3*T) + (3*n + m)*T*e^(-3*T) Recall that T = t - k, and replace accordingly: y(T) = n*e^(-3*(t - k)) + (3*n + m)*(t - k)*e^(-3*(t - k))
If there are multiple dependent variables, x, and y, then Y(s) and X(s) would be two separate Laplace transforms that don't necessarily have anything to do with one another, until a constraint is established. An example of when you would see this, is in systems of diffEq's. Consider the following system, with both x(0) = 0, and y(0) = 0. x'(t) = -6*x(t) + 4*y(t) - 2 y'(t) = -x(t) - 2*y(t) + 5 Take the Laplace of each equation: s*X(s) = -6*X(s) + 4*Y(s) - 2/s s*Y(s) = -X(s) - 2*Y(s) + 5/s Gather X(s) and Y(s) to the LHS, and keep remaining terms on the right. s*X(s) + 6*X(s) - 4*Y(s) = -2/s s*Y(s) + X(s) + 2*Y(s) = 5/s Factor: (s + 6)*X(s) - 4*Y(s) = -2/s X(s) + (s + 2)*Y(s) = 5/s Use Cramer's rule to solve for X(s) & Y(s) Main determinant D = (s + 6)*(s + 2) + 4 = s^2 + 8*s + 16 = (s + 4)^2 Determinant for X: Dx = -2/s*(s + 2) + 4*5/s = 16/s - 2 Determinant for Y: Dy = (s + 6)*5/s + 3/s = 32/s + 5 Construct solution for X(s) & Y(s) X(s) = Dx/D = (48/s - 6)/(s + 4)^2 = (16 - 2*s)/(s*(s + 4)^2) Y(s) = Dy/D = (96/s + 15)/(s + 4)^2 = (32 + 5*s)/(s*(s + 4)^2) Partial fractions for X & Y: X(s) = A/s + B/(s + 4)^2 + C/(s + 4) = 1/s - 6/(s + 4)^2 - 1/(s + 4) Y(s) = D/s + E/(s + 4)^2 + F/(s + 4) = 2/s - 3/(s + 4)^2 - 2/(s + 4) Inverse Laplace for final solution: x(t) = 1 - 6*t*e^(-4*t) - e^(-4*t) y(t) = 2 - 3*t*e^(-4*t) - 2*e^(-4*t)
The most you could do, is use placeholder constants, like defining u = y(0), and v = y'(0). Then, you solve it in terms of u and v as placeholders for these numbers. At the end of the solution of a 2nd order system, two of the numbers will depend on u and v, and the remaining terms (if they exist) will have their own coefficients that are independent of u and v. The terms that depend on u and v, can ultimately have their coefficients replaced with arbitrary coefficients. As an example, consider: y" + 6*y' + 9*y = 1 - e^(-t) let u = y(0) and v = y'(0) Take the Laplace transform: s^2*Y(s) - v - s*u + 6*s*Y(s) - 6*u + 9*Y(s) = 1/s - 1/(s + 1) Shuffle initial conditions to the right, and factor the left: (s^2 + 6*s + 9)*Y(s) = 1/s - 1/(s + 1) + (s + 6)*u + v Isolate Y(s), and factor: Y(s) = (1/s - 1/(s + 1) + (s + 6)*u + v)/(s + 3)^2 Shuffle denominators below, expand and gather: Y(s) = (s^3*u + (7*u + v)*s^2 + (6*u + v)*s + 1)/(s*(s+1)*(s+3)^2) Partial fractions: Y(s) = A/s + B/(s + 1) + C/(s+3)^2 + D/(s+3) A = 1/9, B = -1/4, C = 3*u + v + 1/6, D =1/72*(450*u + 54*v + 37) Since C and D both are functions of u and v, we can leave them as unspecified constants, and get the general solution: Y(s) = 1/9/s - 1/4/(s+1) + C/(s + 3)^2 + D/(s + 3) Inverse Laplace: y(t) = 1/9 - 1/4*e^(-t) + C*t*e^(-3*t) + D*e^(-3*t)
You can make up placeholder initial conditions. For instance, let y(0) = u, and let y'(0) = v. Given: y" + 6*y' + 9*y = 0, and the above placeholder initial conditions. Let Y(s) = £{y(t)} £{y"(t)} = s^2*Y(s) - v - s*u £{y'(t)} = s*Y(s) - u Compile, expand, gather, and shuffle initial conditions to the right:: (s^2 + 6*s + 9)*Y(s) = u*s + v + 6*u Isolate Y(s) and factor: Y(s) = (u*s + v + 6*u)/(s + 3)^2 Arrange the numerator so one part looks like (s +3)/(s + 3)^2: u*s + v + 6*u = u*(s + 3) - 3*u + v + 6*u = u*(s + 3) + v + 3*u Thus: Y(s) = u/(s + 3) + (v + 3*u)/(s + 3)^2 Take inverse Laplace: y(t) = u*e^(-3*t) + (v + 3*u)*t *e^(-3*t) If you just want the general solution, replace u and (v + 3*u) with your two arbitrary constants, such that y(t) = A*e^(-3*t) + B*t*e^(-3*t). If there were a third term due to starting with a non-homogeneous equation, it would have a coefficient that is independent of the initial conditions.
What he's doing, is he's adding zero in a fancy way, so that when he breaks apart the Laplace transform to take its inverse Laplace, he can recognize the components in a standard table of Laplace transforms, and match them to the original function.
Given: y" - 2*y' - 8*y = 0 Assume arbitrary initial conditions of y(0) = u and y'(0) = v, and take the Laplace: s^2*Y(s) - s*u - v - 2*s*Y(s) + 2*u - 8*Y(s) = 0 Shuffle initial conditions to the right, and factor the left: (s^2 - 2*s - 8)*Y(s) = u*s + v - 2*u (s + 2)*(s - 4)*Y(s) = u*s + v - 2*u Isolate Y(s): Y(s) = (u*s + v - 2*u)/((s + 2)*(s - 4)) Partial fractions: Y(s) = A/(s + 2) + B/(s - 4) Since we have two unknowns, and two unspecified initial conditions, we don't really need to solve for A and B. Just leave them as arbitrary constants, since they ultimately both depend on u and v anyway, which are both unknowns. If we had 3 unknowns and 2 initial conditions, then only one of them would be independent of u and v, which would be the one that is part of the particular solution. The coefficients on the homogeneous part of the solution will be the coefficients that depend on initial conditions of u and v. Inverse Laplace, for our general solution of: y(t) = A*e^(-2*t) + B*e^(4*t)
So many of us engineering student is depending on videos like this so thank you so much
: )
true
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this kind of videos MUST be paid!! im grateful its very free and bprp is soooo noice
leopardi the poodle NoName
I am very glad to hear! Best of luck on your finals (if you are taking any soon)
@@blackpenredpen i actually do next week☺
leopardi the poodle NoName
Yup, I guessed it right! Best of luck and let me know how it goes!
Thank you for pointing out the S is not a 5, I make this mistake all the time! these are the tips students need
The best channel about math. Thank you for your videos. It’s really helpful
great video straight to the point. Making calculus more fun and learnable
Your videos are so useful. Congrats for you channel!
God Bless You Bro, Helpful AF!
Thank you for your clean-explanation.
Thank you so much. I used this video in preparation for a national exam and a quetstion similar to this came up. THANKS🙏
You are just a blessing. Period #
you're really helping us lot, we thenk you prof.
You are a true saviour
Great video! I have a question for you: Is there another way to do inverse of Laplace transform? I mean, any formula, theorem... And so on.
Great sir👏
Really good video!
4:48, you can also use the partial fraction decomposition to make it easy!
A real G ! God bless
i like ur vdos
im gonna use this method for my high school de courses. so much better.
why would someone dislike this video ?
Perfect example thank you
Brilliant!
Is this possible without an initial condition? If so can you show it, I'm trying it myself but I get stuck when having to inverse the Laplace transform
You would put in initial conditions as variables. Like letting y(0) equal w, and letting y'(0)=v. Then, you can pick any initial conditions to see the solution in action.
Omg thank you 😭🙈
abigail abigail he's good
Great videos.
but on this you lost me on the last inverse transform ( blue, right hand side). I've never seen this trick of splitting the squared into a 1+1. If someone could explain this or point me in the right direction it would be appreciated.
Laplace{t^n}=n!/(s^[n+1]), so we can deduce that 1/s^2 is t^1.
in my class, the teacher taught us to assume that y=e^(rx) and solve a quadratic equation. If the r doesn't have any real roots, just use cosine and sine.
4:48, you can also use the partial fraction decomposition!
That's one way of solving this kind of diff EQ. There is usually more than one way to solve problems in math, so your way isn't wrong per say. The question could ask you to solve the equation using the Laplace transformation, so you'd need to know how to do this as well.
The Laplace transform is particularly useful when you DON'T have the equation equal to zero, but instead there is a forcing function of t, on the other side of the equals sign.
The assume y=e^(r*t), and solve a quadratic equation works best for the homogeneous cases. He's using this as an example to demonstrate the concept of solving the same problem via a Laplace transform.
Hi great vid, I just have one question regarding the solution, why do you add 3te^3t instead of subtracting it?
Bottom left of the board, he factored the negative out so he could manipulate the insides of the function. When he took the inverse laplace, he distributed the negative again, making the 3 positive.
Hey, I really appreciate your videos. They inspire me a lot !
I also have a request, can you explain the integral of (e^x*cosx) ? It's a partial integral which I don't know how to solve.
I would use the identity cosx= 0.5(e^îx+e^-ix). Then you only have to integrate e-functions.
BouzyWouzy I did the integral of e^xsinx here czcams.com/video/2I-_SV8cwsw/video.html and it should help u with ur problem
Ohh thanks a lot !! It will definitely help me with studying. Like all of your videos, actually :)
Thanks good explanation
Good.
Wow thanks! :)
Thank you
i would get so confused trying to write with two different colored pens in the same hand
laplace transform is straight black magic
Integration by parts is Ultraviolet Voodoo
Thx very very very much
thank you
Great. I have also a channel of mathematics.
Thank you for that.
But I have a question: how to solve an ODE by Laplace when y(k) = n and y'(k) = m, for k =/= 0?
To solve an equation with non-initial conditions instead of initial conditions, you cannot do it directly with the Laplace transform. The Laplace transform only works with initial conditions directly.
However, there are other methods we could use. One method is to assign arbitrary initial conditions at t=0 as placeholders, and then solve for them later to match the given information, such as y(0) = u and y'(0) = v. Another method with the way you've given the information, is to simply assign capital T, such that capital T is zero when t=k. This means that T = t - k.
Then our delayed start conditions, end up being initial conditions after all, and we can proceed with solving the problem in the capital T domain. I'll use capital S to reflect the difference. And then undo the shift, to get back to the little t domain. For this example:
y" + 6*y' + 9*y = 0
y(k) = n, y'(k) = m
S^2*Y(S) - S*n - m + 6*S*Y(S) - 6*n + 9*Y(S) = 0
Shuffle initial conditions to the right:
(S^2 + 6*S + 9)*Y(S) = S*n + 6*n + m
Isolate Y(S) and factor the bottom:
Y(S) = (S*n + 6*n + m)/(S + 3)^2
Partial fractions:
Y(S) = n/(S+3) + (3*n + m)/(S + 3)^2
Solution in capital T world:
y(T) = n*e^(-3*T) + (3*n + m)*T*e^(-3*T)
Recall that T = t - k, and replace accordingly:
y(T) = n*e^(-3*(t - k)) + (3*n + m)*(t - k)*e^(-3*(t - k))
Look at the diifrence! Its only the square!
If y" is equal to s^2 y(s) what now is the equal of 2x"? Is it 2s^2 y(s) thank you in advance
If there are multiple dependent variables, x, and y, then Y(s) and X(s) would be two separate Laplace transforms that don't necessarily have anything to do with one another, until a constraint is established.
An example of when you would see this, is in systems of diffEq's. Consider the following system, with both x(0) = 0, and y(0) = 0.
x'(t) = -6*x(t) + 4*y(t) - 2
y'(t) = -x(t) - 2*y(t) + 5
Take the Laplace of each equation:
s*X(s) = -6*X(s) + 4*Y(s) - 2/s
s*Y(s) = -X(s) - 2*Y(s) + 5/s
Gather X(s) and Y(s) to the LHS, and keep remaining terms on the right.
s*X(s) + 6*X(s) - 4*Y(s) = -2/s
s*Y(s) + X(s) + 2*Y(s) = 5/s
Factor:
(s + 6)*X(s) - 4*Y(s) = -2/s
X(s) + (s + 2)*Y(s) = 5/s
Use Cramer's rule to solve for X(s) & Y(s)
Main determinant D = (s + 6)*(s + 2) + 4 = s^2 + 8*s + 16 = (s + 4)^2
Determinant for X: Dx = -2/s*(s + 2) + 4*5/s = 16/s - 2
Determinant for Y: Dy = (s + 6)*5/s + 3/s = 32/s + 5
Construct solution for X(s) & Y(s)
X(s) = Dx/D = (48/s - 6)/(s + 4)^2 = (16 - 2*s)/(s*(s + 4)^2)
Y(s) = Dy/D = (96/s + 15)/(s + 4)^2 = (32 + 5*s)/(s*(s + 4)^2)
Partial fractions for X & Y:
X(s) = A/s + B/(s + 4)^2 + C/(s + 4) = 1/s - 6/(s + 4)^2 - 1/(s + 4)
Y(s) = D/s + E/(s + 4)^2 + F/(s + 4) = 2/s - 3/(s + 4)^2 - 2/(s + 4)
Inverse Laplace for final solution:
x(t) = 1 - 6*t*e^(-4*t) - e^(-4*t)
y(t) = 2 - 3*t*e^(-4*t) - 2*e^(-4*t)
would it be possible to solve without a boundary condition using laplace ?
This would be an initial value problem, and as far as I'm aware the Laplace transformation only works for IVPs and not BVPs.
@@GhostyOcean fax u cant solve w laplace w out initial vals
The most you could do, is use placeholder constants, like defining u = y(0), and v = y'(0). Then, you solve it in terms of u and v as placeholders for these numbers.
At the end of the solution of a 2nd order system, two of the numbers will depend on u and v, and the remaining terms (if they exist) will have their own coefficients that are independent of u and v. The terms that depend on u and v, can ultimately have their coefficients replaced with arbitrary coefficients.
As an example, consider:
y" + 6*y' + 9*y = 1 - e^(-t)
let u = y(0) and v = y'(0)
Take the Laplace transform:
s^2*Y(s) - v - s*u + 6*s*Y(s) - 6*u + 9*Y(s) = 1/s - 1/(s + 1)
Shuffle initial conditions to the right, and factor the left:
(s^2 + 6*s + 9)*Y(s) = 1/s - 1/(s + 1) + (s + 6)*u + v
Isolate Y(s), and factor:
Y(s) = (1/s - 1/(s + 1) + (s + 6)*u + v)/(s + 3)^2
Shuffle denominators below, expand and gather:
Y(s) = (s^3*u + (7*u + v)*s^2 + (6*u + v)*s + 1)/(s*(s+1)*(s+3)^2)
Partial fractions:
Y(s) = A/s + B/(s + 1) + C/(s+3)^2 + D/(s+3)
A = 1/9, B = -1/4, C = 3*u + v + 1/6, D =1/72*(450*u + 54*v + 37)
Since C and D both are functions of u and v, we can leave them as unspecified constants, and get the general solution:
Y(s) = 1/9/s - 1/4/(s+1) + C/(s + 3)^2 + D/(s + 3)
Inverse Laplace:
y(t) = 1/9 - 1/4*e^(-t) + C*t*e^(-3*t) + D*e^(-3*t)
ur a kingggggg
I love you
thanks!
lmao i was just gonna write this. i also love him. hey its 2017 after all !
Thanks buddy. Why do you need to hold this huge black ball
if he drops it there's an earthquake
It's a microphone.
Is it possible to do this without initial condition?
You can make up placeholder initial conditions.
For instance, let y(0) = u, and let y'(0) = v.
Given: y" + 6*y' + 9*y = 0, and the above placeholder initial conditions.
Let Y(s) = £{y(t)}
£{y"(t)} = s^2*Y(s) - v - s*u
£{y'(t)} = s*Y(s) - u
Compile, expand, gather, and shuffle initial conditions to the right::
(s^2 + 6*s + 9)*Y(s) = u*s + v + 6*u
Isolate Y(s) and factor:
Y(s) = (u*s + v + 6*u)/(s + 3)^2
Arrange the numerator so one part looks like (s +3)/(s + 3)^2:
u*s + v + 6*u = u*(s + 3) - 3*u + v + 6*u = u*(s + 3) + v + 3*u
Thus:
Y(s) = u/(s + 3) + (v + 3*u)/(s + 3)^2
Take inverse Laplace:
y(t) = u*e^(-3*t) + (v + 3*u)*t *e^(-3*t)
If you just want the general solution, replace u and (v + 3*u) with your two arbitrary constants, such that y(t) = A*e^(-3*t) + B*t*e^(-3*t). If there were a third term due to starting with a non-homogeneous equation, it would have a coefficient that is independent of the initial conditions.
What a Chad this guy is
Fourier Series please
what happens if youre not given y'(0)?
SepiaDragoonGR you call it C or whatever you want and treat it as a constant.
in Our Engineering Compus
Y(0)= 1
And Here
Y(0)= -1
Who is Wrong Our Professor or Our Youmather😅😅😅
This is a very common example lol.
Could someone please explain with basic english why he put +3 in the answer.I solved this problem and didn't add +3.
What he's doing, is he's adding zero in a fancy way, so that when he breaks apart the Laplace transform to take its inverse Laplace, he can recognize the components in a standard table of Laplace transforms, and match them to the original function.
you have a mistake: L{y"}=s^2L{y}-sy'-y but in 2nd string you put 6 as y and -1 as y'
Денис Дрожжин no he doesn't
lost me at 5:30
Plz solve this y ^2-2y^1-8y=0
Given:
y" - 2*y' - 8*y = 0
Assume arbitrary initial conditions of y(0) = u and y'(0) = v, and take the Laplace:
s^2*Y(s) - s*u - v - 2*s*Y(s) + 2*u - 8*Y(s) = 0
Shuffle initial conditions to the right, and factor the left:
(s^2 - 2*s - 8)*Y(s) = u*s + v - 2*u
(s + 2)*(s - 4)*Y(s) = u*s + v - 2*u
Isolate Y(s):
Y(s) = (u*s + v - 2*u)/((s + 2)*(s - 4))
Partial fractions:
Y(s) = A/(s + 2) + B/(s - 4)
Since we have two unknowns, and two unspecified initial conditions, we don't really need to solve for A and B. Just leave them as arbitrary constants, since they ultimately both depend on u and v anyway, which are both unknowns. If we had 3 unknowns and 2 initial conditions, then only one of them would be independent of u and v, which would be the one that is part of the particular solution. The coefficients on the homogeneous part of the solution will be the coefficients that depend on initial conditions of u and v.
Inverse Laplace, for our general solution of:
y(t) = A*e^(-2*t) + B*e^(4*t)
hello brother
Cleaner
bRuh, you could've just solved it with a characteristic equation like a normie.
Yeah, but the whole point of the video is to use this method.....
SIR YOU ARE MISTAKEN